Definite IntegrationmediumFree

∫(f−g) for max/min of {x|x|, x²|x|} | JEE

JEE Maths question with a full step-by-step solution.

Question

Let

f(x)

be the maximum and

g(x)

be the minimum of

\big\{x|x|,\,x^{2}|x|\big\}

. Then

\displaystyle\int_{-1}^{1}\big[f(x)-g(x)\big]\,dx

equals

\dfrac{1}{12}

\dfrac{1}{3}

\dfrac{2}{3}

correct

\dfrac{7}{12}

Solution

Step 1: Compare the two functions. Since

x|x|-x^{2}|x|\ge 0\ \Rightarrow\ |x|\big(x-x^{2}\big)\ge 0,

this holds when

x=0,1

x-x^{2}>0\Rightarrow 0<x<1

. So

x|x|\ge x^{2}|x|

[0,1]

, and

x|x|\le x^{2}|x|

for

x<0

x>1

. Step 2: Read off the maximum

f

and minimum

g

on each region:

f(x)=\begin{cases}x|x|,&0\le x\le 1\\[4pt]x^{2}|x|,&x<0\ \text{or}\ x>1\end{cases}\qquad g(x)=\begin{cases}x^{2}|x|,&0\le x\le 1\\[4pt]x|x|,&x<0\ \text{or}\ x>1\end{cases}

Hence

f-g=\big|x|x|-x^{2}|x|\big|

. On

[0,1]

|x|=x

gives

f-g=x^{2}-x^{3}

; on

[-1,0]

|x|=-x

gives

f-g=(x^{3}-x^{2})

taken positive, i.e.

x^{2}-x^{3}

. Step 3: On the whole of

[-1,1]

therefore

f(x)-g(x)=x^{2}-x^{3}

, hence

\int_{-1}^{1}\big(x^{2}-x^{3}\big)\,dx.

Step 4: The term

x^{3}

is odd, so it integrates to

0

;

x^{2}

is even:

\int_{-1}^{1}\big(x^{2}-x^{3}\big)\,dx=2\int_{0}^{1}x^{2}\,dx-0=2\cdot\dfrac13=\dfrac{2}{3}.

Correct answer: (3)

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